# How can I calculate $\mathrm{E}\!\left[\sum _{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\right]$?

Suppose that $Y_1,\dots,Y_{n+1}$ is a random sample from a continuous distribution function $F$. Let$X\sim\mathrm{Uniform}\{1,\dots,n\}$ be independent of the $Y_i$'s. How can I compute $\mathrm{E}\!\left[\sum _{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\right]$?

• This looks like the conditional expectation of a random variable $I$, given that $Y_i \leq Y_{n+1}$. What is $I$? Or, were you trying to write an indicator function? Like $I \{ Y_i \leq Y_{n+1} \}$? Mar 8, 2017 at 18:13
• If $I$ is an indicator function just use linearity of expectation. Mar 8, 2017 at 18:17
• What do those vertical bars mean in "$I\mid Y_i\le Y_{n+1}\mid$"? This is not a conventional notation for an indicator function, which raises doubts concerning what this question is asking.
– whuber
Mar 8, 2017 at 22:39
• Maybe the OP only meant to use the first vertical bar. That could mean that it is intended to mean conditioning. Mar 9, 2017 at 2:49
• @Zen I believe you might have misunderstood: somebody's (your?) edits had fixed the notational problem, not created them! With the rollback, the strange notation has returned.
– whuber
Mar 9, 2017 at 16:18

Here is an alternative answer to @Lucas' using the law of iterated expectations:

\begin{align} E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}\right] & = E\left[E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}|X\right]\right] \\ & = E\left[\sum_{i=1}^XE[1_{(Y_i \leq Y_{n+1})}|X]\right] \\ & = E\left[\sum_{i=1}^XE[1_{(Y_i \leq Y_{n+1})}]\right] \\ & = E\left[\sum_{i=1}^XE\left[E[1_{(Y_i \leq Y_{n+1})}|Y_{n+1}]\right]\right] \\ & = E\left[\sum_{i=1}^XE[F(Y_{n+1})]\right] \\[12pt] & = E[X]E\left[F(Y_{n+1})\right] \\[12pt] & =\frac{n+1}{2}E[F(Y_{n+1})] \end {align}

The third step follows from independence of $Y_i$ and $Y_{n+1}$ from $X$; the fourth step is again an application of the law of iterated expectations; the last step is simply an application of the formula for the expectation of a discrete uniform random variable.

By inverting the order of integration, we derive the remaining expectation:

\begin{align} E[F(Y_{n+1})] & = \int_{-\infty}^{\infty}F(y)dF(y) \\ & = \int_{-\infty}^{\infty} \int_{-\infty}^y dF(x)dF(y) \\ & = \int_{-\infty}^{\infty} \int_{x}^{\infty} dF(y)dF(x) \\ & = \int_{-\infty}^{\infty} (1-F(x))dF(x) \\[10pt] & = 1-E[F(Y_{n+1})] \end{align}

which implies $E[F(Y_{n+1})] = \frac{1}{2}$. Hence:

$$E\left[\sum_{i=1}^X1_{(Y_i \leq Y_{n+1})}\right] = \frac{n+1}{4}$$

By distributional symmetry, $\Pr\{Y_i\leq Y_{n+1}\}=\Pr\{Y_{n+1}\leq Y_i\}$, for each $i=1,\dots,n$. Since $F$ is continuous, we have $$\Pr\{Y_i\leq Y_{n+1}\} = 1-\Pr\{Y_{n+1}< Y_i\}=1-\Pr\{Y_{n+1}\leq Y_i\}.$$ Therefore, $\mathrm{E}\left[I_{\{Y_i\leq Y_{n+1}\}}\right]=\Pr\{Y_i\leq Y_{n+1}\}=1/2$. Now, we have $$\mathrm{E}\!\left[\sum_{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X=x\right] = \mathrm{E}\!\left[\sum_{i=1}^x I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X=x\right] = \sum_{i=1}^x\;\mathrm{E}\!\left[I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X=x\right]$$ $$= \sum_{i=1}^x\;\mathrm{E}\!\left[I_{\{Y_i\leq Y_{n+1}\}}\right] = \frac{x}{2},$$ in which we used the linearity of the conditional expectation and the independence of $X$ and the $Y_i$'s. Hence, $$\mathrm{E}\!\left[\sum_{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\right] = \mathrm{E}\!\left[\mathrm{E}\!\left[\sum_{i=1}^X I_{\{Y_i\leq Y_{n+1}\}}\;\Bigg\vert\; X\right]\right] = \mathrm{E}\left[\frac{X}{2}\right] = \frac{n+1}{4}.$$

We have \begin{align} E\left[ \sum_{i = 1}^X I[Y_i \leq Y_{n + 1}] \right] &= E\left[ \sum_{i = 1}^n I[i \leq X] I[Y_i \leq Y_{n + 1}] \right] \\ &= \sum_{i = 1}^n E\left[ I[i \leq X] I[Y_i \leq Y_{n + 1}] \right] \\ &= \sum_{i = 1}^n E\left[ I[i \leq X] \right] \cdot E\left[ I[Y_i \leq Y_{n + 1}] \right] \\ &= \sum_{i = 1}^n \frac{i}{n} \cdot E[I[Y_i \leq Y_{n + 1}]] \\ &= \sum_{i = 1}^n \frac{i}{n} \cdot E\left[ F(Y_{n + 1})] \right] \\ &= \sum_{i = 1}^n \frac{i}{n} \cdot \frac{1}{2} \\ &= \frac{n + 1}{4} \end{align}

The second step follows from the linearity of expectations, the third step from the independence of $X$ and $Y_1, ..., Y_{n + 1}$, and the fifth step from the fact that $$F(y) = P(Y \leq y) = E[I[Y \leq y]].$$ To prove the sixth step, you can use partial integration. For the final step, you use the formula for partial sums.

• Where does your $N$ come from? As long as $N>n-1$, that is always going to be true hence you can fix the final result at the value you want. Mar 8, 2017 at 20:48
• If you mean: $N=n$ I do agree with the answer. Mar 8, 2017 at 21:22